An engine operating on a different cycle mustnecessarily transfer heat across finite temperaturedifference and therefore cannot be reversible.

Carnot Cycle with an Ideal Gas as Working Fluid

Step1: a-

b

Reversible

adiabatic compression until thetemperature rise from TC

to TH

Step 2: b–

c

Reversibleisothermal expansion to arbitrarypoint c with heat absorption of QH

Step 3: c–

d

Reversible

adiabatic expansion until thetemperature decrease TC.

Step 4: d–

a

Reversible

isothermal compression to the initial state with heat rejection of QC

Entropy

Our analysis showed that

or

The equation suggests the existence of a property whose changes are given bythe quantities Q/T

When theisotermal

steps are infinitesimal, the heat quantities becomedQ:

or

Thus the quantitiesdQrev/T sum to zero for the arbitrary cycle, exhibiting thecharacteristic of a property. We call this property as entropy, S.

0TQTQCCHHCHCHTTQQ0TdQTdQCCHH0TdQrevTdQdSrevdSTdQrevThere exists a property called entropy S, which is an intrinsic property of asystem, functionally related to the measurable coordinates whichcharacterize the system

Characteristic of Entropy

Entropy owes its existence to the second law, from which itarises in much the same way as internal energy does from thefirst law.

The change in entropy of any system undergoing a finitereversible

process is

When a system undergoes anirreversible

process betweentwo states, the entropy change of the system is evaluated to anarbitrary chosen reversibleprocess that accomplishes thesame change of state as the actual process. The integration isnot

carried out for the irreversible path.

TdQSrevCharacteristic of Entropy

Since entropy is a state function, the entropy changes of theirreversible and reversible processes are identical.

The entropy change of a system caused by the transfer of heatcan always be calculated bydQ/T whether the heat transferis accomplished reversibly or irreversibly.

When a process is irreversible on account of finite differencesin other driving forces, such as pressure, the entropy changesis not caused solely by heat transfer, and for its calculationsone must devise a reversible means of accomplishing the samechange of state.

Calculation of Entropy

For one mole of fluid undergoing a mechanically reversibleprocess in a closed system, using the first law, the definingequation for enthalpy one find:

For an ideal gas:

Although derived for a mechanically reversible process, thisequation is a general equation for the calculation of entropychanges, since it relates properties only and independent ofprocess causing the change of state.

TdPVTdHTdQdSrev0TTigpPPlnTdTRCRS0Example

CalculateΔS for each step of the cycle shown below.Assume ideal gas with constant Cp.

Problem 5.6

A quantity of an ideal gas, Cp

= 7/2 R at 20oC & 1 barand having a volume of 70 m3

is heated at constantpressure to 25oC by the transfer of heat from a heatreservoir at 40oC. Calculate the heat transfer to thegas, the entropy change of the heat reservoir, theentropy change of the gas, andStotal.

Problem 5.7

A rigid vessel of 0.05 m3

volume contains an ideal gas,Cv

= 5/2 R, at 500 K and 1 bar.

a.If heat in the amount of 12000 J is transferred tothe gas, determine its entropy change

b.If the vessel is fitted with a stirrer that is rotated bya shaft so that work on the amount of 12000 J isdone on the gas, what is the entropy change of thegas if the process is adiabatic? What isStotal?

Sketch this cycle on a PV diagram and determine its thermalefficiency if T1

= 500 K, T2

= 800 K, T3

= 2000 K, and T4

=1000 K

Problem 5.13

A reversible cycle executed by 1 mol of an ideal gas for whichCp

= 5/2 R consist of the following processes:

a.Starting at 600 K & 2 bar, the gas is cooled at constantpressure to 300 K

b.From 300 K & 2 bar, the gas is compressed isothermallyto 4 bar

c.The gas returns to its initial state along a path for whichthe product PT is constant.

What is the thermal efficiency of the cycle?

Example 5.4

A 40-kg steel casting (Cp = 0.5 kJ/(kg.K) at atemperature of 450oC is quenched in 150 kg of oil(Cp = 2.5 kJ/(kg.K) at 25oC. If there are no heatlosses, what is the change of entropy of:

a.The casting

b.The oil

c.Both considered together

Problem 5.11

A piston/cylinder device contains 5 mol of an ideal gas,Cp

= 5/2 R, at 20oC & 1 bar. The gas is compressedreversibly and adiabatically to 10 bar, where thepiston is locked in position. The cylinder is thenbrought into thermal contact with a heat reservoir at20oC, and heat transfer continues until the gas alsoreaches this temperature. Determine the entropychanges of the gas, the reservoir, andStotal

Irreversibility

One mole of an ideal gas, Cp = 7/2 R is compressedadiabatically in a piston/cylinder device from 2 bar &25oC to 7 bar. The process is irreversible and requires35% more work than a reversible adiabaticcompression from the same initial state to the samefinal pressure. What is the entropy change of thegas?

Classical Lost Work & Process Analysis

The ideal work is the maximum amount of work whichcan be done by the process by operating reversiblywithin the system and by transferring heat betweenthe system and the surroundings reversibly.

The lost work:

For processes containing several units, lost-workcalculations can be made for each unit and summedto determine the overall value.

Determine temperature, entropy change of the steamleaving the turbine, & the lost work

Example 4.8

Assume that 5000 kg/h of oil with a heat capacity of3.2 kJ/kg.K

is to be cooled from 220 to 40oC, using alarge quantity of water which can be assumed to beat a constant temperature of 30oC. Determine thelost work in the process and the thermodynamicefficiency of the process.

Example 4.9

Assume that 100 kg of methane gas/h is adiabaticallycompressed from 0.5MPa

& 300 K to 3.0MPa

&500 K after which it is cooledisobarically

to 300 Kby a large amount of water available at 290 K.Determine the efficiency of the compressor.

If the surroundings are assumed to be at 290 K, do athermodynamic analysis of the process.

Assume ideal gas.

Cp = 35.58 J/mol.K

Cv

= 27.27 J/mol.K

8

Steamdengan

tekanan

12 bardan

temperatur

200oC

diekpansi

satu

tahap

dalam

sebuah

turbin

sehingga

tekanannya

menjadi

1,5 barpada

kondisi

jenuhnya.Turbin

bekerja

secara

adiabatik

dengan

efisiensi

85%.Perkirakan

(dalam

sistem

satuan

SI):

Kualitas

uap

keluar

turbin

Kebutuhan

uap

air agardihasilkan

daya

sebesar

50MWatt

Kerja

ideal,kerja

musnah

dan

perubahan

entropi

total,jika

temperatur

lingkungan

25oC

Problem 5.8

An ideal gas, Cp

= 7/2 R, is heated in a steady-flow heatexchanger from 20oC to 100oC by another stream ofthe same ideal gas which enters at 180oC. The flowrates of the two streams are the same, and heatlosses from the exchanger are negligible.

a.Calculate the molar entropy changes of the two gasstreams for both parallel and counter-current flowin the exchanger